Check sibling questions

Example 47 - Find intervals in which f(x) = 3/10x4 - 4/5x3

Example 47 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 47 - Chapter 6 Class 12 Application of Derivatives - Part 3
Example 47 - Chapter 6 Class 12 Application of Derivatives - Part 4


Transcript

Example 47 Find intervals in which the function given by f(π‘₯) =3/10 π‘₯4 – 4/5 π‘₯^3– 3π‘₯2 + 36/5 π‘₯ + 11 is (a) strictly increasing (b) strictly decreasingf(π‘₯) = 3/10 π‘₯4 – 4/5 π‘₯^3– 3π‘₯2 + 36/5 π‘₯ + 11 Finding f’(𝒙) f’(π‘₯) = 3/10 Γ— 4π‘₯^3 – 4/5 Γ— 3π‘₯^2 – 3 Γ— 2x + 36/5 + 0 f’(π‘₯) = 12/10 π‘₯^3– 12/5 π‘₯^2– 6x + 36/5 f’(π‘₯) = 6/5 π‘₯^3βˆ’ 12/5 π‘₯^2– 6x + 36/5 f’(π‘₯) = 6(π‘₯^3/5βˆ’(2π‘₯^2)/5βˆ’π‘₯+6/5) f’(π‘₯) = 6((π‘₯^3 βˆ’ 2π‘₯^2βˆ’ 5π‘₯ + 6)/5) = 6/5 (π‘₯^3βˆ’2π‘₯^2βˆ’5π‘₯+6) = 6/5 (π‘₯βˆ’1)(π‘₯2βˆ’π‘₯βˆ’6) = 6/5 (π‘₯βˆ’1)(π‘₯2βˆ’3π‘₯+2π‘₯βˆ’6) = 6/5 (π‘₯βˆ’1)[π‘₯(π‘₯βˆ’3)+2(π‘₯βˆ’3)] = 6/5 (π‘₯βˆ’1)(π‘₯+2)(π‘₯βˆ’3) Hence, f’(𝒙) = πŸ”/πŸ“ (π’™βˆ’πŸ)(𝒙+𝟐)(π’™βˆ’πŸ‘) Putting f’(𝒙) = 0 πŸ”/πŸ“ (π’™βˆ’πŸ)(𝒙+𝟐)(π’™βˆ’πŸ‘) = 0 (π‘₯βˆ’1)(π‘₯+2)(π‘₯βˆ’3) = 0 Hence, x = –2 , 1 & 3 Plotting points on number line Hence, f(π‘₯) is strictly decreasing on the interval π‘₯ ∈ (βˆ’βˆž,βˆ’πŸ)& (𝟏 , πŸ‘) f(π‘₯) is strictly increasing on the interval π‘₯ ∈ (βˆ’πŸ,𝟏) & (πŸ‘ , ∞)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.